# Ex 5.3, 4 - Chapter 5 Class 10 Arithmetic Progressions

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Ex 5.3 ,4 How many terms of the AP. 9, 17, 25 โฆ must be taken to give a sum of 636? AP is 9,17,25, โฆ.. We know that Sum = ๐/2 (2a + (n โ 1) d) Here, a = 9 d = 17 โ 9 = 8 & Sum = Sn = 636 We need to find n Putting values in equation Sum = ๐/2 (2a + (n โ 1) d) 636 = ๐/(2 ) (2 ร 9 + (n โ 1) ร 8) 636 ร 2 = n (18 + (n โ 1) ร 8) 1272 = n (18 + 8n โ 8) 1272 = n (10 + 8n) 1272 = 10n + 8n2 0 = 10n + 8n2 โ 1272 10n + 8n2 โ 1252 = 0 2(5n + 4n2 โ 636)= 0 5n + 4n2 โ 636 = 0 4n2 + 5n โ 636 = 0 We solve by quadratic formula Comparing equation with ax2 + bx + c = 0 Here, a = 4, b = 5, c = โ 636 We know that, D = b2 โ 4ac D = (5)2 โ 4 ร 4 ร (โ636) D = 25 + 10176 D = 10201 Hence, roots to equation are given by n = (โ ๐ ยฑ โ๐ท)/2๐ Putting values n = (โ 5 ยฑ โ10201)/(2 ร 4) n = (โ 5 ยฑ 101)/8 Solving So, n = 12 & n = (โ53)/4 Since n cannot be negative, n = 12

Chapter 5 Class 10 Arithmetic Progressions

Serial order wise

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